Question

A lift with open top is moving up with an acceleration $f$. A body is thrown up with speed $u$ relative to lift and it comes back to lift in time. Then the speed $u$ will be given as, $\left( g=\text{acceleration due to gravity} \right)$
$\begin{align}
  & A.ft+gt \\
 & B.\sqrt{{{\left( ft \right)}^{2}}+{{\left( gt \right)}^{2}}} \\
 & C.\dfrac{ft+gt}{2} \\
 & D.\dfrac{{{f}^{2}}t}{g} \\
\end{align}$

Answer 1

Hint: First of all we have to find the net acceleration, which will be the sum of acceleration of the lift and the acceleration due to gravity. Using these details, find the time of flight of the lift, which is the ratio of twice the speed to the net acceleration. Rearranging this equation in terms of speed will give the required solution.

Complete answer:
In the lift, the object inside the lift is acted by the acceleration of the lift as well as the acceleration due to gravity. Therefore the net acceleration is to be found here.

The acceleration of the lift is given as,
$a=f$
And acceleration due to gravity is $g$.
This can be written in an equation as,
${{a}_{net}}=a+g=f+g$
The time of flight is given by the equation,
$t=\dfrac{2u}{f+g}$
Where the time of flight is indicated as $t$, the net acceleration is given as $f+g$, and $u$ be the speed of the object inside the lift. We have to find the speed of the object inside the lift. Therefore the equation is to be rearranged in terms of the speed,
$\begin{align}
  & 2u=ft+gt \\
 & u=\dfrac{ft+gt}{2} \\
\end{align}$

Therefore the correct answer is given as option C.

Note:
Time of flight is the measurement of the time taken by an object, particle or wave to travel a distance through a medium. Speed is a scalar quantity. That is the quantity is dependent only on the magnitude of the quantity. The acceleration due to gravity is the gravitational acceleration provided by the earth.